Find the x and y Intercepts y=3x^2+6x+4

$y = 3 x^{2} + 6 x + 4$

Step 1

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = 3 x^{2} + 6 x + 4$

Solve the equation.

Rewrite the equation as $3 x^{2} + 6 x + 4 = 0$ .

$3 x^{2} + 6 x + 4 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 3$ , $b = 6$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (3 \cdot 4)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 3 \cdot 4}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 4$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 6 \pm \sqrt{36 - 12 \cdot 4}}{2 \cdot 3}$

Multiply $- 12$ by $4$ .

$x = \frac{- 6 \pm \sqrt{36 - 48}}{2 \cdot 3}$

Subtract $48$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 12}}{2 \cdot 3}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 3}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 6 \pm i \cdot \sqrt{12}}{2 \cdot 3}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 6 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 3}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 6 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm i \cdot (2 \sqrt{3})}{2 \cdot 3}$

Move $2$ to the left of $i$ .

$x = \frac{- 6 \pm 2 i \sqrt{3}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 6 \pm 2 i \sqrt{3}}{6}$

Simplify $\frac{- 6 \pm 2 i \sqrt{3}}{6}$ .

$x = \frac{- 3 \pm i \sqrt{3}}{3}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 3 \cdot 4}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 4$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 6 \pm \sqrt{36 - 12 \cdot 4}}{2 \cdot 3}$

Multiply $- 12$ by $4$ .

$x = \frac{- 6 \pm \sqrt{36 - 48}}{2 \cdot 3}$

Subtract $48$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 12}}{2 \cdot 3}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 3}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 6 \pm i \cdot \sqrt{12}}{2 \cdot 3}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 6 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 3}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 6 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm i \cdot (2 \sqrt{3})}{2 \cdot 3}$

Move $2$ to the left of $i$ .

$x = \frac{- 6 \pm 2 i \sqrt{3}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 6 \pm 2 i \sqrt{3}}{6}$

Simplify $\frac{- 6 \pm 2 i \sqrt{3}}{6}$ .

$x = \frac{- 3 \pm i \sqrt{3}}{3}$

Change the $\pm$ to $+$ .

$x = \frac{- 3 + i \sqrt{3}}{3}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 + i \sqrt{3}}{3}$

Factor $- 1$ out of $i \sqrt{3}$ .

$x = \frac{- 1 \cdot 3 - (- i \sqrt{3})}{3}$

Factor $- 1$ out of $- 1 (3) - (- i \sqrt{3})$ .

$x = \frac{- 1 (3 - i \sqrt{3})}{3}$

Move the negative in front of the fraction.

$x = - \frac{3 - i \sqrt{3}}{3}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 3 \cdot 4}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 4$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 6 \pm \sqrt{36 - 12 \cdot 4}}{2 \cdot 3}$

Multiply $- 12$ by $4$ .

$x = \frac{- 6 \pm \sqrt{36 - 48}}{2 \cdot 3}$

Subtract $48$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 12}}{2 \cdot 3}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 3}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 6 \pm i \cdot \sqrt{12}}{2 \cdot 3}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 6 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 3}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 6 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm i \cdot (2 \sqrt{3})}{2 \cdot 3}$

Move $2$ to the left of $i$ .

$x = \frac{- 6 \pm 2 i \sqrt{3}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 6 \pm 2 i \sqrt{3}}{6}$

Simplify $\frac{- 6 \pm 2 i \sqrt{3}}{6}$ .

$x = \frac{- 3 \pm i \sqrt{3}}{3}$

Change the $\pm$ to $-$ .

$x = \frac{- 3 - i \sqrt{3}}{3}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 - i \sqrt{3}}{3}$

Factor $- 1$ out of $- i \sqrt{3}$ .

$x = \frac{- 1 \cdot 3 - (i \sqrt{3})}{3}$

Factor $- 1$ out of $- 1 (3) - (i \sqrt{3})$ .

$x = \frac{- 1 (3 + i \sqrt{3})}{3}$

Move the negative in front of the fraction.

$x = - \frac{3 + i \sqrt{3}}{3}$

The final answer is the combination of both solutions.

$x = - \frac{3 - i \sqrt{3}}{3}, - \frac{3 + i \sqrt{3}}{3}$

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

x-intercept(s): $None$

Step 2

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = 3 {(0)}^{2} + 6 (0) + 4$

Solve the equation.

Remove parentheses.

$y = 3 \cdot 0^{2} + 6 (0) + 4$

Remove parentheses.

$y = 3 {(0)}^{2} + 6 (0) + 4$

Simplify $3 {(0)}^{2} + 6 (0) + 4$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 3 \cdot 0 + 6 (0) + 4$

Multiply $3$ by $0$ .

$y = 0 + 6 (0) + 4$

Multiply $6$ by $0$ .

$y = 0 + 0 + 4$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 4$

Add $0$ and $4$ .

$y = 4$

y-intercept(s) in point form.

y-intercept(s): $(0, 4)$

Step 3

List the intersections.

x-intercept(s): $None$

y-intercept(s): $(0, 4)$

Step 4

Find the x and y Intercepts y=3x^2+6x+4

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